JUST THE MATHS SLIDES NUMBER 8.2 VECTORS 2 (Vectors in component - PDF document

“JUST THE MATHS” SLIDES NUMBER 8.2 VECTORS 2 (Vectors in component form) by A.J.Hobson 8.2.1 The components of a vector 8.2.2 The magnitude of a vector in component form 8.2.3 The sum and difference of vectors in component form 8.2.4 The direction cosines of a vector

UNIT 8.2 - VECTORS 2 VECTORS IN COMPONENT FORM 8.2.1 THE COMPONENTS OF A VECTOR Vectors in space are defined in terms of unit vectors placed along the axes O x , O y and O z of a three-dimensional right-handed cartesian reference system. These unit vectors will be denoted respectively by i , j , and k ; (“bars” and “hats” may be omitted). Consider the following diagram: z ✻ P( x, y, z ) � � r � C � ✻ � B � ✲ ✲ S y ✟ ✟ ✟ O ❅✟✟✟✟✟✟✟✟✟✟✟ A ✟ ✟ ✙ ❅ ❅ ✟ ✟ ❅ ❅ ✟ ✟ ❅ ❅ ✟ ✟ ❅ ❅ ✟ ✟ ❅ ❅ R ✟ ❅ ✟ ✟ Q( x, y, 0) ✟ ✙ ✟ x In the diagram, OA = i , OB = j and OC = k . P is the point with co-ordinates ( x, y, z ). 1

By the Triangle Law, r = OP = OQ + QP = OR + RQ + QP . That is, r = x i + y j + z k . Note: Vectors which emanate from the origin are not a special case since we are dealing with free vectors. Nevertheless, OP is called the position vector of the point P. The numbers x , y and z are called the “components” of OP (or of any other vector in space with the same magnitude and direction as OP). 2

8.2.2 THE MAGNITUDE OF A VECTOR IN COMPONENT FORM z ✻ P( x, y, z ) � � r � C � ✻ � B � ✲ ✲ S y ✟ ✟ O ✟ ❅✟✟✟✟✟✟✟✟✟✟✟ A ✟ ✟ ✙ ❅ ❅ ✟ ✟ ❅ ❅ ✟ ✟ ❅ ❅ ✟ ✟ ❅ ❅ ✟ ✟ ❅ ❅ R ✟ ❅ ✟ ✟ Q( x, y, 0) ✟ ✟ ✙ x By Pythagoras’ Theorem, (OP) 2 = (OQ) 2 + (QP) 2 = (OR) 2 + (RQ) 2 + (QP) 2 That is, � x 2 + y 2 + z 2 . r = | x i + y j + z k | = EXAMPLE Determine the magnitude of the vector a = 5 i − 2 j + k and hence obtain a unit vector in the same direction. 3

Solution √ � 5 2 + ( − 2) 2 + 1 2 = | a | = a = 30 . A unit vector in the same direction as a is obtained by normalising a, that is, dividing it by its own magnitude. The required unit vector is a = 1 a . a = 5 i − 2 j + k √ . � 30 8.2.3 THE SUM AND DIFFERENCE OF VECTORS IN COMPONENT FORM Consider, first, a situation in two dimensions: y ✻ ✏ Q ✏✏✏✏✏✏✏✏ ✁ ✁ P 1 ✁ ✁ ✁ ✕ ✁ ✁ ✁ ✁ ✏ ✶ ✏✏✏✏✏✏✏✏ ✁ S P 2 ✁ ✁ ✲ x O R In the diagram, suppose P 1 has co-ordinates ( x 1 , y 1 ) and suppose P 2 has co-ordinates ( x 2 , y 2 ). ∆ gl ORP 1 has exactly the same shape as ∆ gl P 2 SQ. Hence, the co-ordinates of Q must be ( x 1 + x 2 , y 1 + y 2 ). 4

By the Parallelogram Law, OQ is the sum of OP 1 and OP 2 . That is, ( x 1 i + y 1 j ) + ( x 2 i + y 2 j ) = ( x 1 + x 2 ) i + ( y 1 + y 2 ) j . It can be shown that this result applies in three dimensions also. To find the difference of two vectors, we calculate the difference of their separate components. 5

EXAMPLE Two points P and Q in space have cartesian co-ordinates ( − 3 , 1 , 4) and (2 , − 2 , 5) respectively. Determine the vector PQ. Solution Q ✟ ✟✟✟✟ ✄ ✯ ✟ ✟✟ ✄ P ✄ ❆ ❆ ✄ ✄✄ ✗ ❆ ✄ ❆ ✄ ✄ ❑ ❆ ❆ ❆ ✄ ✄ ❆ ❆ ✄ ✄ O OP = − 3 i + j + 4 k . OQ = 2 i − 2 j + 5 k . By the triangle Law, PQ = OQ − OP = 5 i − 3 j + k . Note: The vector drawn from the origin to the point (5 , − 3 , 1) is the same as the vector PQ. 6

8.2.4 THE DIRECTION COSINES OF A VECTOR Suppose that OP = r = x i + y j + z k and suppose that OP makes angles α , β and γ with O x , O y and O z respectively. Then cos α = x cos β = y r and cos γ = z r , r . z ✻ ✁ ✕ ✁ ✁ γ ✁ ✁ ✁ ✲ y ✟ O ✟ ✟ ✟ ✟ x ✙ ✟ cos α , cos β and cos γ are called the “direction cosines” of r. Any three numbers in the same ratio as the direction cosines are said to form a set of “direction ratios” for the vector r. x : y : z is one possible set of direction ratios. 7

EXAMPLE The direction cosines of the vector 6 i + 2 j − k are 6 2 − 1 √ √ √ 41 and 41 , 41 since the vector has magnitude √ 36 + 4 + 1 = √ 41. A set of direction ratios for this vector are 6 : 2 : − 1. 8